Two charges 7 and -2 are placed at (-9, 0, 0) cm and (9, 0, 0) cm respectively in an external field E = Ar2 r, where A = 9 105 N m2 C-1. Considering the potential at infinity is 0, the electrostatic energy of the configuration is J.
- 49.3
- 1.4
- 24.3
- -90.7
Answer
(1) 49.3
Solution
Potential of the external field (V = 0 at infinity): V(r) = - r Ar2 dr = Ar
Both charges are at r = 9 cm = 0.09 m V = 9 1050.09 = 107 V
Energy in the external field: q1 V + q2 V = (7 - 2) 10-6 107 = 50 J
Mutual energy (r12 = 0.18 m): k q1 q2r12 = 9 109 7 10-6 (-2 10-6)0.18 = -0.7 J
U = 50 - 0.7 = 49.3 J
Answer: (1)